poj2400

题解：

最少平均分值是等于最佳匹配的权值和除上一个总的点数2*n

注意输入反过来

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>  
const int N=21;  
int n,lx[N],ly[N],T,w[N][N],k,d[N],t,ans,num,h[N],px[N],py[N];  
int dfs1(int x)
{  
    px[x]=1;  
    for (int i=1;i<=n;i++)
     {  
        if (py[i])continue;  
        t=lx[x]+ly[i]-w[x][i];  
        if (!t)
         {  
            py[i]=1;  
            if (!d[i]||dfs1(d[i]))
             {  
                d[i]=x;  
                return 1;  
             }  
         }  
     }   
    return 0;  
}  
void dfs2(int x,int sum)
{   
    if (sum>ans) return;  
    if (x>n)
     {  
        if (sum!=ans)return;  
        printf("Best Pairing %d
",++num);  
        for (int i=1;i<=n;i++)  
         printf("Supervisor %d with Employee %d
",i,h[i]);  
        return;  
     }  
    for (int i=1;i<=n;i++)  
     if (!py[i])
      {  
        h[x]=i;  
        py[i]=1;  
        dfs2(x+1,sum-w[x][i]);  
        py[i]=0;  
      }  
    return;  
}  
int main()
{  
    scanf("%d",&T);  
    for (int u=1;u<=T;u++)
     {  
        scanf("%d",&n);  
        for (int i=1;i<=n;i++)  
         for (int j=1;j<=n;j++)
          {  
            scanf("%d",&k);  
            w[k][i]=1-j;  
          }  
        for (int i=1;i<=n;i++)  
         for (int j=1;j<=n;j++)
          {  
            scanf("%d",&k);  
            w[i][k]+=1-j;  
          }  
        memset(ly,0,sizeof(ly));  
        for (int i=1;i<=n;i++)
         {  
            lx[i]=w[i][1];  
            for (int j=2;j<=n;j++)  
             if (lx[i]<w[i][j])lx[i]=w[i][j];  
         }  
        memset(d,0,sizeof(d));  
        for (int i=1;i<=n;i++)  
         while (1)
          {  
            memset(px,false,sizeof(px));  
            memset(py,false,sizeof(py));  
            if (dfs1(i)) break;  
            t=1e9;  
            for (int j=1;j<=n;j++)  
             if (px[j])  
              for (int k=1;k<=n;k++)  
               if (!py[k]&&lx[j]+ly[k]-w[j][k]<t)  
                t=lx[j]+ly[k]-w[j][k];  
            for (int j=1;j<=n;j++)
             {  
                if (px[j]) lx[j]-=t;  
                if (py[j]) ly[j]+=t;  
             }  
          }  
        ans=0;  
        for (int i=1;i<=n;i++)  
         if (d[i]!=0)ans-=w[d[i]][i];  
        printf("Data Set %d, Best average difference: %.6f
",u,ans*0.5/n);  
        num=0;  
        memset(h,0,sizeof(h));  
        memset(py,false,sizeof(py));  
        dfs2(1,0);  
        if (u<T)puts("");  
     }   
    return 0;  
}